How to Determine a Power Series Represents an Exponential Function Suppose I define the function 
$$
f(x) = \sum_{n = 0}^\infty \frac{x^n}{n!}.
$$
Is there anything I can directly observe about the power series to conclude that it is the expansion of an exponential function?
Noting that $f^\prime(x) = f(x)$ and concluding $f(x) = e^x$ is not quite what I have in mind. Rather, I'm hoping there is some observation that would first allow me to conclude $f(x) = a^x$ for some $a > 0$. Having discovered a very special base for an exponential, I would then define $e = \sum_{n=0}^\infty \frac{1}{n!}$.
 A: It's not too hard to prove that $f(x+y) = f(x) f(y)$ in a "nice" way, which implies that $f$ is an exponential function.
The binomial theorem says $$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k} = \sum_{i+j=n} \frac{n!}{i!\,j!} x^i y^j$$ where I've chosen the second form to be more symmetric with respect to $x$ and $y$. (The sum, to be clear, is over all pairs $(i,j)$ with $i+j=n$.)
Therefore $$f(x+y) = \sum_{n=0}^\infty \frac{(x+y)^n}{n!} = \sum_{n=0}^\infty \left(\sum_{i+j=n} \frac{1}{i!\,j!}x^i y^j\right).$$
Summing over all $n$ and then over all $i$ and $j$ that add up to $n$ is the same as summing over all $i$ and all $j$ to begin with, so this is equal to
$$\sum_{i=0}^\infty \sum_{j=0}^\infty \frac{x^i y^j}{i!\,j!} = \left(\sum_{i=0}^\infty \frac{x^i}{i!}\right) \left(\sum_{j=0}^\infty \frac{x^j}{j!}\right) = f(x) f(y).$$
This seems like just an unmotivated algebraic expansion, but the combinatorial underpinnings of this are actually quite important: this is exactly the same justification that we use to explain what multiplying together two exponential generating functions does to the objects they're counting. (For the purposes of understanding $f$, of course, that's irrelevant.)
A: You say that $f'=f$ is not what you have in mind. But this leads to $f'(x)/f(x)=1$, so
$$
x=\int_0^x\frac{f'(t)}{f(t)}\,dt=\int_1^{f(x)}\frac1u\,du.
$$
The above shows that the function 
$$
g(x)=\int_1^x\frac1u\,du
$$
is the inverse of $f$. Playing with this function one finds that $g(1)=0$, and with $v=u/y$, 
\begin{align}
g(xy)&=\int_1^{xy}\frac1u\,du=\int_{1/y}^x\frac{y}{yv}\,dv=\int_1^x\frac1v\,dv+\int_{1/y}^1\frac1v\,dv\\ \ \\
&=\int_1^x\frac1v\,dv+\int_{1}^y\frac1u\,du
=g(x)+g(y).
\end{align}
Then
$$
f(x+y)=f(g(f(x))+g(f(y)))=f(g(f(x)f(y))=f(x)f(y).
$$
In particular, for $n\in\mathbb N$,
$$\tag{1}
f(nx)=f(x)^n.
$$
Also,
$$\tag{2}
f(1)=f(n\,\frac1n)=f(1/n)^n,
$$
so $f(1/n)=f(1)^{1/n}$. Combining $(1)$ and $(2)$ we get, for every $q\in\mathbb Q$,
$$
f(q)=f(1)^q.
$$
At this point, we might as well define the number $e$ to be $f(1)$. Since $f$ is continuous (because the series converges uniformly) it makes sense to define arbitrary powers of $f(1)$ by 
$$
f(1)^x=f(x).
$$
So $f(x)=e^x$. 
A: You can directly show that the power series for $f$ satisfies $f(x+y) = f(x)f(y)$.  Note that the coefficient of $x^iy^j$ on both sides equals $\frac{1}{i!j!} = \frac{1}{(i+j)!}\binom{i+j}{i}$.  This property is saying that $f$ is a homomorphism from the additive reals onto the multiplicative group of nonzero reals.
It is straightforward to check by induction that this guarantees that $f(n) = f(1)^n$ and $f(1/n) = f(1)^{1/n}$.  More generally, we have that $f(q) = f(1)^q$ for all rational numbers $q$.  Since $f(x)$ is continuous, it therefore must be equal to $f(1)^x$.
A: We get that series from a Taylor series expansion of $e^x$. Taylor series expansions are a separate topic and $e^x$ just happens to be a special case.
It's just another continuous, infinitely differentiable function: Wikipedia article on Taylor Series.
I suppose to answer your question, the power series is exponential because of how Taylor series expansions are defined. For an infinitely differentiable $f$, the expansion about $a$
$$\sum_{n=0}^{\infty} = \frac{f^{(n)}(a)}{n!} (x-a)^n $$
is exponential regardless of $f$. 
If your question, on the other hand, is if someone "randomly" came up the series in your question, how would you find out they have in fact defined $e^x$ near $0$, then I believe that is a completely different question. However, just judging by the title and description, I'm not sure if that's what you meant.
